6.4. Factorization and property (T) 231Remark 6.4.9. It is possible to avoid Lemma 6.2.5 in the proof above, as
follows. Write
h = P1Q 1 E9 P2Q 2 E9 ... E9 PkQk
q q q
where ~^1 < P: < · · · < P; are the nonzero eigenvalues of hand Qi, ... , Qk
are the corresponding spectral projections. Define a *-homomorphismby
1f^1 (x) = Qicr(x)Q1 E9 · · · E9 QkCJ(x)Qk.
There is a trace T on QiIIB(1-t'.)Q1 E9 · · · E9 QkIIB(H)Qk such that T( 1f^1 (x)) =
'Tr(CJ(x)h) (just think about it) and hence we can apply Exercise 6.2.3 and
replace the finite-dimensional algebra QiIIB(1-t'.)Q1 E9 · · · E9 QkIIB(H)Qk with a
full matrix algebra.The following corollary is immediate from the previous result and The-
orem 6.2.7.
Corollary 6.4.10. Assumer has property (T) and T is a tracial state on
C (r). Then T is amenable if and only if there exist -homomorphisms
7rn: C(f)-+ Mk(n)(q such that limtr(7rn(x)) = T(x) for all x E C(r).
Remark 6.4.11. In most instances, the previous corollary is what we need,
but note that more is true. Namely, if J <1 C(r) is an ideal and T is an
amenable trace on the quotient C(f)/ J, then there exist -homomorphisms
7rn: C(r)/J-+ Mk(n)(C)
such that lim tr(7rn(x)) = T(x) for all x E C(f)/ J. This is because Propo-
sition 6.4.8 also holds for C(f)/ J (the proof amounts to changing a bit of
notation).
We are now very close to Kirchberg's result asserting property (T) groups
with the factorization property are residually finite; we just need an impor-
tant theorem of Malcev on the structure of finitely generated linear groups.
Theorem 6.4.12. Let K be a field which is finitely generated as a ring.
Then K is finite.
Though elementary and well known, this result doesn't appear in many
basic algebra texts; hence we include the proof at the end of this section.
Deducing Malcev's result from it, however, is easy.
Theorem 6.4.13 (Malcev). Let K be a field. Every finitely generated sub-
group of GL(n, K) is residually finite.